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Math Help - Eigenvalues

  1. #1
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    Eigenvalues

    M is a nxn matrix where all entries are 1's.

    My conjecture is that n - 1 eigenvalues are 0 and the nth eigenvalues is one.

    If this is correct, how do I go about proving it?
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  2. #2
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    Re: Eigenvalues

    did you test your conjecture for small values of n? for a 2x2 matrix:

    \begin{vmatrix}1-x&1\\1&1-x\end{vmatrix} = (1-x^2) - 1 = 1 - 2x + x^2 - 1 = x^2 - 2x = x(x-2)

    which appears to have eigenvalues 0 and 2.
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  3. #3
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    Re: Eigenvalues

    Quote Originally Posted by Deveno View Post
    did you test your conjecture for small values of n? for a 2x2 matrix:

    \begin{vmatrix}1-x&1\\1&1-x\end{vmatrix} = (1-x^2) - 1 = 1 - 2x + x^2 - 1 = x^2 - 2x = x(x-2)

    which appears to have eigenvalues 0 and 2.
    Now I am not sure what it may be.
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  4. #4
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    Re: Eigenvalues

    try a 3x3 matrix next, maybe you'll get a theory?
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    Re: Eigenvalues

    Using the steps:

    (i)\;R_i\to R_i-R_1\;(i>1)\qquad (ii)\;C_1\to C_1+C_2+\ldots C_n

    we obtain

    \displaystyle\begin{aligned}\begin{vmatrix} 1-\lambda & 1 & \ldots & 1\\ 1 &1-\lambda & \ldots & 1 \\ \vdots&&&\vdots \\ 1 & 1 &\ldots & 1-\lambda\end{vmatrix}&=\begin{vmatrix} 1-\lambda & \;\;1 & \ldots & \;\;1\\ \lambda &-\lambda & \ldots & \;\;0 \\ \vdots&&&\vdots \\ \lambda & \;\;0 &\ldots & -\lambda\end{vmatrix}\\&=\begin{vmatrix} n-\lambda & 1 & \ldots & 1\\ 0 &-\lambda & \ldots & 0 \\ \vdots&&&\vdots \\ 0 & 0 &\ldots & -\lambda\end{vmatrix}\\&=(n-\lambda)(-\lambda)^{n-1}\end{aligned}

    So, the eigenvalues are \lambda=0 (n-1-uple) and \lambda=n (1-uple) if n\geq 2 . If n=1 we have only the eigenvalue \lambda=1 (1-uple).
    Last edited by FernandoRevilla; October 14th 2011 at 10:43 PM.
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  6. #6
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    Re: Eigenvalues

    i would have been inclined use induction and expand along the top row, but that works great!
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    MHF Contributor FernandoRevilla's Avatar
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    Re: Eigenvalues

    In general with the steps

    (i)\;R_i\to R_i-R_1\;(i>1)\qquad (ii)\;C_1\to C_1+C_2+\ldots C_n

    we triangularize the determinant

    \begin{vmatrix} a & b & \ldots & b\\ b & a & \ldots & b \\ \vdots&&&\vdots \\ b & b &\ldots & a\end{vmatrix}=\ldots=[a+(n-1)b](a-b)^{n-1}
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  8. #8
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    Re: Eigenvalues

    Quote Originally Posted by FernandoRevilla View Post
    Using the steps:

    (i)\;R_i\to R_i-R_1\;(i>1)\qquad (ii)\;C_1\to C_1+C_2+\ldots C_n

    we obtain

    \displaystyle\begin{aligned}\begin{vmatrix} 1-\lambda & 1 & \ldots & 1\\ 1 &1-\lambda & \ldots & 1 \\ \vdots&&&\vdots \\ 1 & 1 &\ldots & 1-\lambda\end{vmatrix}&=\begin{vmatrix} 1-\lambda & \;\;1 & \ldots & \;\;1\\ \lambda &-\lambda & \ldots & \;\;0 \\ \vdots&&&\vdots \\ \lambda & \;\;0 &\ldots & -\lambda\end{vmatrix}\\&=\begin{vmatrix} n-\lambda & 1 & \ldots & 1\\ 0 &-\lambda & \ldots & 0 \\ \vdots&&&\vdots \\ 0 & 0 &\ldots & -\lambda\end{vmatrix}\\&=(n-\lambda)(-\lambda)^{n-1}\end{aligned}

    So, the eigenvalues are \lambda=0 (n-1-uple) and \lambda=n (1-uple) if n\geq 2 . If n=1 we have only the eigenvalue \lambda=1 (1-uple).
    I don't understand how you obtained n - \lambda in the a_{11} position.
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  9. #9
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    Re: Eigenvalues

    he started with 1-λ, and added (n-1) 1's. 1-λ + (n-1) = n-λ.
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    Re: Eigenvalues

    Quote Originally Posted by Deveno View Post
    he started with 1-λ, and added (n-1) 1's. 1-λ + (n-1) = n-λ.
    How did he get the first column to be all zero after n-lambda? Why did only that entry change? How can we just add something to a_{11} and leave everything else alone?
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  11. #11
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    Re: Eigenvalues

    first he subtracted the first row, from every other row. this changed all the 1's in the first column (except for the top entry) to -λ.

    it also changed all the 1's in every other column to 0, except along the diagonal, where 1-λ becomes -λ.

    then he sucessively added (it doesn't matter if you do it one at a time, or all at once) columns 2 through n, to the first column.

    neither the row operation, nor the column operation changes the determinant, since no row or column was multiplied.
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  12. #12
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    Re: Eigenvalues

    Quote Originally Posted by Deveno View Post
    first he subtracted the first row, from every other row. this changed all the 1's in the first column (except for the top entry) to -λ.

    it also changed all the 1's in every other column to 0, except along the diagonal, where 1-λ becomes -λ.

    then he sucessively added (it doesn't matter if you do it one at a time, or all at once) columns 2 through n, to the first column.

    neither the row operation, nor the column operation changes the determinant, since no row or column was multiplied.
    Ok it was the adding of the columns I couldn't figure out.
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  13. #13
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    Re: Eigenvalues

    Now the eigenvectors.

    For \lambda =0, I obtained

    \left\{\begin{bmatrix}-1\\n\\0\\ \vdots\\0\end{bmatrix},\begin{bmatrix}-1\\0\\n\\ \vdots\\0\end{bmatrix},\cdots,\begin{bmatrix}-1\\0\\0\\ \vdots\\n\end{bmatrix}\right\}

    However, I am finding it hard to solve for lambda = n

    Is it just

    \begin{bmatrix}1\\0\\0\\ \vdots\\0\end{bmatrix}
    Last edited by dwsmith; October 15th 2011 at 04:33 PM.
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  14. #14
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    Re: Eigenvalues

    the eigenvalue n has a multiplicity of 1, so all we need to do is find ONE eigenvector. try (1,1,1,.....,1).

    (how did i find this? note that (A-nI)(x1,x2,....,xn) = (0,0,....,0) implies that

    x1+x2+....+xn = nxj for ALL j)
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  15. #15
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    Re: Eigenvalues

    Quote Originally Posted by Deveno View Post
    the eigenvalue n has a multiplicity of 1, so all we need to do is find ONE eigenvector. try (1,1,1,.....,1).

    (how did i find this? note that (A-nI)(x1,x2,....,xn) = (0,0,....,0) implies that

    x1+x2+....+xn = nxj for ALL j)
    Shouldn't we get this:

    \begin{vmatrix}0&1&\cdots & 1\\0&-n&\cdots &0\\ \vdots & 0& \ddots \\ 0&\cdots & &-n \end{vmatrix}

    x_2=-x_3-x_4-\cdots -x_n

    x_i=0 \ \ i=2,...,n

    x_1 is a free variable.

    So shouldn't it just be the n dimensional e_1 vector?
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